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# (17*19*23*29)^31 = n. Lowering which of the following number

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Senior Manager
Joined: 20 May 2014
Posts: 282
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Kudos [?]: 49 [0], given: 220

(17*19*23*29)^31 = n. Lowering which of the following number [#permalink]  12 Nov 2017, 00:53
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$$(17*19*23*29)^{31} = n$$. Lowering which of the following numbers by one will result in the least decrease of n?

A. 17
B. 19
C. 23
D. 29
E. 31

Kudos for correct solution.
[Reveal] Spoiler: OA
Director
Joined: 03 Sep 2017
Posts: 521
Followers: 1

Kudos [?]: 327 [2] , given: 66

Re: (17*19*23*29)^31 = n. Lowering which of the following number [#permalink]  13 Nov 2017, 00:17
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This question is about a pattern meaning that we have not to solve the expression given but we can make up an easier one since the rule is general.

Let's try with $$(1*2*3*4)^2 = 24^2 = 576$$. It is easy to notice that if we reduce the exponent we would get the largest decrease, from 576 to 24.

Then we can focus on the components of the product, if try to reduce each one by 1, we find that the smallest decrease is found when we reduce 4, the largest integer.

Applying this finding to the original expression, 29 is the number, if decreased, which decreases the least the expression.

Re: (17*19*23*29)^31 = n. Lowering which of the following number   [#permalink] 13 Nov 2017, 00:17
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